Where does this equation derive from?

Question

I saw this equation yesterday, and was wondering where it came from.

Range = (Vo² sin(2theta))/a
Where range is the change in x
Vo is the initial velocity
and a is the acceleration

Thanks.

Answer 1

Vox = initial velocity in the x direction
Voy = initial velocity in the y direction
g = acceleration due to gravity

Range = Vox * t = Vo*sin(theta)*t
y = (g/2)*t² + Voy * t = (g/2)*t^2 + Vo *cos(theta) * t
we want y to be 0 (so the object is at to the same height it started at) so
0 = (g/2)*t² + Vo *cos(theta) * t
solve for t
0 = (g/2)*t² + Vo *cos(theta) * t
divide by t
0 = (g/2)*t + Vo *cos(theta)
(g/2)*t = -Vo * cos(theta)
t = -2 * Vo * cos(theta)/a
plug that into Range = Vo*sin(theta)*t
Range = Vo*sin(theta)*(-2 * Vo * cos(theta)/a)
Range = Vo² * -2 * sin(theta) * cos(theta)/a
Use the trig identity 2*sin(u)*cos(u) = sin(2u)
Range = Vo² * -sin(2theta) / g
g = -a (you use a positive acceleration and gravity is negitive so we make a = -g to make it positive)
and we get:

Range = Vo² * -sin(2theta) / -a
= Vo² * sin(2theta) / a

Answer 2

Briefly:

The velocity vector can be broken into its horizontal and vertical components. The horizontal component stays constant (since there is no force in the horizontal direction), and is given by:

Vx = VoCosθ

So the range R is just this horizontal velocity multiplied by the time the projectile stays in the air:

Range = Vx·T= VoCosθ·T

So now you just need to find the value of "T" (total time in the air). To do that, you look at the vertical component of the velocity. It is not constant, but changes as a function of time, due to gravity. Specifically, the change in vertical velocity is: ΔVy = gt, where "t" is how long it's been in the air so far.

Byt the time the object has gone up and come back down again, its change in vertical velocity is twice its initial vertical velocity. That means, after the full time T has elapsed:

ΔVy = gT = 2Vy_initial = 2VoSinθ

T = 2VoSinθ/g

Combined with the previous equation for Range, this gives:

Range = VoCosθ·2VoSinθ/g
= (Vo² sin(2θ))/g

Answer 3

might have something to do with range of a projectile? are you talking about classical mechanise and equations of motion?